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Question 1

Topic: 3.3

(a) The movement of molecules within an organism can occur by diffusion and active transport.

Complete Table 1.1 by placing ticks (✓) to show the correct features of each process.

(b) Explain why active transport is important in root hair cells.

▶️ Answer/Explanation

Solution

(a)

Explanation: Active transport requires energy (✓) as it moves substances against their concentration gradient, while diffusion is passive (no ✓ for energy). Diffusion is responsible for gas exchange (✓) and its rate increases with a steeper concentration gradient (✓). Both processes occur across membranes, but the question specifies “always” for active transport (✓) since diffusion can also occur in open spaces.

(b)

Active transport is crucial in root hair cells because:

It allows absorption of mineral ions even when their concentration in the soil is lower than inside the root cells
It enables movement of ions against their concentration gradient, which is essential for nutrient uptake
It helps create a water potential gradient that facilitates water absorption by osmosis
Specific ions like nitrates and magnesium are often in low concentration in soil but needed in large amounts by plants

Detailed Explanation: Root hair cells are specialized for absorbing water and minerals from the soil. The concentration of essential mineral ions in the soil water is usually much lower than their concentration inside the root hair cells. Diffusion alone couldn’t move these ions into the roots because it only works along the concentration gradient. Active transport uses energy from respiration to pump these ions into the root against their concentration gradient. This process not only provides the plant with necessary nutrients but also lowers the water potential inside the root, causing water to follow by osmosis. Without active transport, plants would struggle to obtain enough minerals, especially in nutrient-poor soils.

Question 2

Topic: 8.3

Transpiration is the loss of water vapour from the leaves of a plant.

(a) Complete the sentence describing transpiration.

Water evaporates from the surfaces of the …… cells into the air spaces and then …… out of the leaves, through the stomata, as water vapour.

(b) Explain how water moves upwards in the xylem.

They measured the rate of transpiration when the plant was placed in different wind speeds in both the light and the dark. The tree was given an adequate supply of water.

The results of this investigation are shown in Fig. 2.1.

(i) Describe and explain the results shown in Fig. 2.1.

(ii) The investigation was repeated in the light with the same species of tree in an environment where the humidity was higher.

Draw a line on Fig. 2.1 to predict the results in the light with increased humidity.

(iii) The investigation described in 2(c) was repeated (different wind speeds in both the light and the dark).

However, the tree was not given an adequate supply of water during the investigation.

Predict the result and explain your prediction.

▶️ Answer/Explanation

Solution

(a) mesophyll; diffuses;

Explanation: Transpiration begins when water evaporates from the moist surfaces of mesophyll cells (the photosynthetic cells in leaves) into the air spaces within the leaf. This water vapor then diffuses out through the stomata, which are small pores on the leaf surface that allow for gas exchange.

(b)

Water moves upwards in the xylem due to three main factors:

Transpiration pull: As water evaporates from the leaves (transpiration), it creates a negative pressure that pulls water up through the xylem vessels.
Cohesion-tension theory: Water molecules stick together (cohesion) due to hydrogen bonding, forming a continuous column of water from roots to leaves.
Adhesion: Water molecules also adhere to the walls of the xylem vessels, helping to counter gravity.

This combined action creates a continuous flow of water from roots to leaves, known as the transpiration stream.

(c)(i)

Description:

As wind speed increases, the rate of transpiration increases in both light and dark conditions.
Transpiration rates are consistently higher in light conditions compared to dark conditions at all wind speeds.
The rate of increase is more pronounced in light conditions, especially at higher wind speeds (above 4 m/s).

Explanation:

Increased wind speed removes water vapor from around the leaf faster, maintaining a steeper concentration gradient for diffusion.
In light, stomata are open for photosynthesis, allowing more water vapor to escape, while in dark conditions many stomata close.
Light also provides energy that increases the kinetic energy of water molecules, enhancing evaporation.

(c)(ii)

A line should be drawn that shows transpiration increasing with wind speed but remaining below the original light condition line at all points.

Explanation: Higher humidity means the air already contains more water vapor, reducing the concentration gradient between the leaf interior and exterior. This decreases the rate of transpiration at all wind speeds compared to normal humidity conditions.

(c)(iii)

Prediction: The transpiration rate would be lower at all wind speeds, particularly in light conditions, with little change in dark conditions.

Explanation:

With inadequate water supply, the plant conserves water by closing stomata (especially in light when they would normally be open).
Closed stomata dramatically reduce water loss, lowering transpiration rates.
In dark conditions, where stomata are already mostly closed, the effect would be less noticeable.
The plant may wilt as cells lose turgor pressure due to water shortage.

Question 3

Topic: 9.2

Fig. 3.1 is a diagram of a section through a human heart.

(a) Draw an X on the septum in Fig. 3.1.

(b) Explain the reason for the difference between the thickness of the walls at B and at C in Fig. 3.1.

(c) A red blood cell enters the vena cava at A in Fig. 3.1. Explain how the red blood cell is moved from the vena cava to the aorta.

(d) An athlete measured her heart rate during a running race. She recorded it before the race, during the race and during her recovery. Her results are shown in Fig. 3.2.

(i) Suggest how the athlete could monitor the activity of her heart.

(ii) Calculate the percentage change in heart rate from the maximum heart rate reached during the race until the heart rate recorded at 18 minutes. Give your answer to three significant figures.

(iii) Explain why heart rate must increase during exercise.

▶️ Answer/Explanation

Solution

(a)

Answer: The X should be drawn on the wall separating the left and right sides of the heart (the septum).

Explanation: The septum is the muscular wall that divides the heart into left and right sides, preventing the mixing of oxygenated and deoxygenated blood.

(b)

Answer: The wall at B (left ventricle) is thicker than at C (right ventricle) because the left ventricle pumps blood at higher pressure to the whole body, while the right ventricle only pumps blood to the lungs.

Explanation: The left ventricle needs more muscular tissue to generate the force required to pump blood throughout the entire body. The right ventricle has a thinner wall because it only needs to pump blood to the nearby lungs, which requires less pressure. This structural difference reflects their different functional demands.

(c)

Answer: The red blood cell moves: from vena cava → right atrium → right ventricle → pulmonary artery → lungs → pulmonary veins → left atrium → left ventricle → aorta.

Explanation: The journey begins when deoxygenated blood enters the right atrium through the vena cava. When the right atrium contracts, it pushes the blood through the tricuspid valve into the right ventricle. The right ventricle then contracts, sending blood through the pulmonary valve into the pulmonary artery to the lungs. In the lungs, the blood becomes oxygenated and returns via the pulmonary veins to the left atrium. The left atrium contracts, pushing blood through the mitral valve into the left ventricle. Finally, the powerful left ventricle contracts, sending oxygenated blood through the aortic valve into the aorta for distribution throughout the body. Valves prevent backflow at each stage.

(d)(i)

Answer: By using an ECG, measuring pulse rate, or listening to heart sounds with a stethoscope.

Explanation: Modern athletes might use smartwatches or fitness trackers that measure heart rate continuously. Alternatively, they could manually measure pulse at the wrist or neck, counting beats per minute. A doctor might use a stethoscope to listen to valve closures.

(d)(ii)

Answer: -48.9%

Explanation: From the graph: Maximum heart rate = 180 bpm at about 5 minutes; Heart rate at 18 minutes = 92 bpm. Percentage change = [(92 – 180)/180] × 100 = (-88/180) × 100 = -48.888…% ≈ -48.9% (to 3 significant figures). The negative sign indicates a decrease in heart rate during recovery.

(d)(iii)

Answer: Heart rate increases to deliver more oxygen and glucose to muscles and remove carbon dioxide and lactic acid faster.

Explanation: During exercise, muscles work harder and require more energy. The increased heart rate serves several crucial functions: 1) It delivers more oxygen to muscles for aerobic respiration, 2) It supplies more glucose (fuel for respiration), 3) It removes carbon dioxide (a waste product of respiration) more quickly, 4) It transports lactic acid (produced during anaerobic respiration) to the liver for processing, and 5) It helps regulate body temperature by increasing blood flow to the skin. This coordinated response ensures muscles can continue contracting efficiently during physical activity.

Question 4

Topic: 13.1

(a) Urea is a waste product.

(i) Describe how urea is formed.

(ii) State the component of blood that transports urea.

(iii) State why urea must be excreted.

(iv) State the name of the blood vessel that carries blood away from the kidney.

(b) Fig. 4.1 shows a drawing of a nephron in the human kidney and associated blood vessels.

Describe how the structures labelled in Fig. 4.1 produce urine.

Explain the importance of nitrate ions to plants.

(d) A farmer applied fertiliser to a field next to a lake.

Suggest two precautions the farmer should take when applying fertiliser to reduce the risk of eutrophication occurring in the lake.

▶️ Answer/Explanation

Solution

(a)(i) Urea is formed through deamination in the liver. This process involves the removal of the nitrogen-containing part of excess amino acids, which are converted into ammonia. Ammonia is then combined with carbon dioxide to form urea, a less toxic compound that can be safely transported in the blood and excreted by the kidneys.

(a)(ii) Plasma.

Explanation: Urea is dissolved in the plasma, the liquid component of blood, which transports it to the kidneys for excretion.

(a)(iii) Urea must be excreted because it is toxic and can harm the body if allowed to accumulate.

Explanation: High levels of urea can disrupt cellular functions and lead to serious health issues, so it must be removed from the body.

(a)(iv) Renal vein.

Explanation: The renal vein carries filtered blood, now with reduced urea levels, away from the kidney and back into the circulatory system.

(b) The glomerulus (K) filters blood under high pressure, allowing water, glucose, urea, and ions to pass into the nephron (L). The nephron then reabsorbs all glucose and some water and ions back into the blood through the capillary network (M). The remaining fluid, now called urine, flows through the collecting duct (N) and contains urea, excess water, and excess ions.

Explanation: This process ensures that essential substances are retained while waste products are efficiently removed from the body.

(c) Nitrate ions are essential for plants because they are a key component in the synthesis of amino acids, which are the building blocks of proteins. Proteins are vital for growth, repair, and various metabolic processes in plants.

Explanation: Without nitrate ions, plants would be unable to produce sufficient proteins, leading to stunted growth and poor development.

(d) Two precautions the farmer should take are:

Apply the fertiliser in appropriate amounts to avoid excess runoff into the lake.
Avoid applying fertiliser before heavy rain to prevent it from being washed into the lake.

Explanation: These precautions help minimise the risk of eutrophication, which occurs when excess nutrients in water bodies lead to algal blooms, oxygen depletion, and harm to aquatic life.

Question 5

Topic: 16.4

(a) Describe the role of progesterone in pregnancy.

(b) Fig. 5.1 shows a drawing of a section of a human placenta. The arrows show the direction of blood flow.

Substances move across the placenta by diffusion.

(i) State the names of two useful substances that move from the blood of the mother to the blood of the fetus.

(ii) Using the information in Fig. 5.1, suggest how the placenta is adapted for efficient diffusion.

(c) The Rhesus (Rh) factor is a protein that is found on the surface of red blood cells in some people. If the protein is present then the individual is Rh positive.

The allele for the presence of the Rh factor is dominant and is represented by the letter D. The recessive allele is represented by the letter d.

If a mother is Rh negative and the fetus is Rh positive there can be problems during pregnancy.

A Rh negative mother and a heterozygous Rh positive father have a child.

Complete the Punnett square and determine the probability of the child being Rh positive.

▶️ Answer/Explanation

Solution

(a)

Progesterone plays several crucial roles in maintaining pregnancy:

It maintains the thickness of the uterine lining (endometrium) to support the developing embryo
It prevents the menstrual cycle from continuing during pregnancy by inhibiting the secretion of FSH and LH
It inhibits contractions of the uterine muscles, preventing premature labor
It prevents the development of additional follicles and ovulation during pregnancy

Without sufficient progesterone, the uterine lining would break down, leading to a miscarriage.

(b)(i)

Two useful substances that move from mother to fetus:

Oxygen – essential for fetal respiration and energy production
Glucose – provides energy for fetal growth and development

Other acceptable answers include amino acids, water, ions, and antibodies.

(b)(ii)

The placenta is adapted for efficient diffusion through:

Large surface area: The numerous villi and microvilli increase the area available for exchange
Short diffusion distance: The placental membrane is very thin, allowing rapid diffusion
Maintained concentration gradient: The constant flow of maternal and fetal blood ensures substances continue to diffuse
Good blood supply: Both maternal and fetal sides have extensive capillary networks

(c)

Punnett square:

Probability of child being Rh positive (having at least one D allele): 50% or 0.5 or ½

Explanation: The mother can only pass on ‘d’ alleles (as she’s Rh negative), while the father can pass either ‘D’ or ‘d’ (as he’s heterozygous). This results in two possible Rh positive (Dd) and two Rh negative (dd) combinations out of four, giving a 50% chance.

Question 6

Topic: 20.4

(a) Fig. 6.1 shows the percentage of land covered by forest in a country from 1660 until 2000.

(i) Using the information in Fig. 6.1, state which 10-year period had the highest increase in the percentage of land covered by forest.

(ii) Outline how human activities could cause the change in the percentage of land covered by forest from 1660 to 1800 shown in Fig. 6.1.

(b) It is estimated that 40% of plant species are at risk of extinction. Seed banks can store seeds from many species for a long time. Table 6.1 shows some information about one plant species that is stored in a seed bank.

(i) Using the data shown in Table 6.1, calculate the mass of protein in one seed.

(ii) Suggest why many plants can easily be conserved using seed banks.

(iii) Describe why conservation projects such as seed banks are important.

▶️ Answer/Explanation

Solution

(a)(i) 1950s

Explanation: By examining the graph in Fig. 6.1, we can see that the steepest increase in forest coverage occurred during the 1950s decade. This period shows the most dramatic upward trend compared to other decades, indicating the highest increase in percentage of land covered by forest.

(a)(ii) Human activities that could cause changes in forest coverage from 1660 to 1800:

1. Deforestation for agriculture: As human populations grew during this period, more land was cleared for farming and livestock grazing, reducing forest coverage.

2. Urbanization: The expansion of towns and cities required clearing forests for building materials and space for development.

3. Industrial use: Forests were cut down for fuel (firewood, charcoal) and raw materials for early industries like shipbuilding and construction.

4. Timber trade: Increased demand for wood products led to commercial logging operations that removed large areas of forest.

(b)(i) 0.2156 g

Calculation: Mass of protein = Total mass of seed × (Percentage protein/100) = 1.96 g × (11/100) = 1.96 × 0.11 = 0.2156 g

(b)(ii) Reasons why plants can be easily conserved in seed banks:

1. Small size: Seeds are compact and require minimal storage space compared to mature plants, allowing many species to be stored in a small area.

2. Long-term viability: Many seeds can remain dormant for years or even decades when stored under proper conditions (low temperature and humidity).

3. Low maintenance: Once properly stored, seeds require little attention or resources to maintain their viability.

(b)(iii) Importance of conservation projects like seed banks:

1. Biodiversity preservation: Seed banks help maintain genetic diversity by storing seeds from various plant species, preventing their extinction.

2. Future research: Preserved seeds provide material for scientific studies on plant genetics, evolution, and potential medicinal uses.

3. Food security: They safeguard crop varieties that may be needed for future food production, especially as climate changes affect agriculture.

4. Ecosystem restoration: Stored seeds can be used to reintroduce native plants to areas where they’ve been lost, helping restore damaged ecosystems.

5. Education: Seed banks serve as valuable resources for educating people about plant conservation and the importance of biodiversity.

Question 7

Topic: 5.1

(a) State the name of a chemical element that is found in all proteins but not in carbohydrates or fats.

(b) Table 7.1 shows the names of some biological molecules, enzymes and the organs that produce the enzymes.

Complete Table 7.1.

▶️ Answer/Explanation

Solution

(a) nitrogen

Explanation: Proteins are unique among these macromolecules because they contain nitrogen atoms in their amino acid building blocks. Carbohydrates and fats are composed of carbon, hydrogen, and oxygen only, while proteins additionally contain nitrogen (and sometimes sulfur). This nitrogen is crucial for forming peptide bonds between amino acids and for creating the diverse structures of proteins.

(b)

Explanation: The table shows the complete breakdown pathways for different macromolecules. Oils (lipids) are broken down by lipase into fatty acids and glycerol. Glycogen is broken down by glycogen phosphorylase into glucose units. The original table had an error where protein was incorrectly associated with maltose production – this actually applies to starch breakdown by amylase. The producing organs are specific to each enzyme’s location in the digestive system.

(c) insulin and glucagon

Explanation: These two pancreatic hormones work antagonistically to regulate glycogen levels in the liver. Insulin promotes glycogen synthesis (glycogenesis) when blood glucose levels are high, storing excess glucose as glycogen. Glucagon triggers glycogen breakdown (glycogenolysis) when blood glucose levels drop, releasing glucose into the bloodstream. Adrenaline can also affect glycogen levels during stress responses, but the primary regulators are insulin and glucagon.

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